assignment on buffer solution

14.6 Buffers

Learning objectives.

By the end of this section, you will be able to:

• Describe the composition and function of acid–base buffers
• Calculate the pH of a buffer before and after the addition of added acid or base

A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer . Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added ( Figure 14.14 ). A solution of acetic acid and sodium acetate (CH 3 COOH + CH 3 COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH 3 ( aq ) + NH 4 Cl( aq )).

How Buffers Work

To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, strong base added to this solution will neutralize hydronium ion, causing the acetic acid ionization equilibrium to shift to the right and generate additional amounts of the weak conjugate base (acetate ion):

Likewise, strong acid added to this buffer solution will shift the above ionization equilibrium left, producing additional amounts of the weak conjugate acid (acetic acid). Figure 14.15 provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.

Example 14.20

Ph changes in buffered and unbuffered solutions.

(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.

(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.

(a) Following the ICE approach to this equilibrium calculation yields the following:

Substituting the equilibrium concentration terms into the K a expression, assuming x << 0.10, and solving the simplified equation for x yields

(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.

Adding strong base will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.

The initial molar amount of acetic acid is

The amount of acetic acid remaining after some is neutralized by the added base is

The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of

Compute molar concentrations for the two buffer components:

Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).

The amount of hydronium ion initially present in the solution is

The amount of hydroxide ion added to the solution is

The added hydroxide will neutralize hydronium ion via the reaction

The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).

The amount of hydroxide ion remaining is

corresponding to a hydroxide molarity of

The pH of the solution is then calculated to be

In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75).

Check Your Learning

Initial pH of 1.8 × × 10 −5 M HCl; pH = −log[H 3 O + ] = −log[1.8 × × 10 −5 ] = 4.74 Moles of H 3 O + in 100 mL 1.8 × × 10 −5 M HCl; 1.8 × × 10 −5 moles/L × × 0.100 L = 1.8 × × 10 −6 Moles of H 3 O + added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L × × 0.0010 L = 1.0 × × 10 −4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant ( Figure 14.16 ). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.

The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.

Selection of Suitable Buffer Mixtures

There are two useful rules of thumb for selecting buffer mixtures:

• Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.

Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H 2 CO 3 , and the bicarbonate ion, HCO 3 − . HCO 3 − . When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:

An added hydroxide ion is removed by the reaction:

The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H 3 O + is converted to H 2 CO 3 and OH - is converted to HCO 3 - ). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.

The Henderson-Hasselbalch Equation

The ionization-constant expression for a solution of a weak acid can be written as:

Rearranging to solve for [H 3 O + ] yields:

Taking the negative logarithm of both sides of this equation gives

which can be written as

where p K a is the negative of the logarithm of the ionization constant of the weak acid (p K a = −log K a ). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation , to calculate the pH of buffer solutions. It is important to note that the “ x is small” assumption must be valid to use this equation.

Portrait of a Chemist

Lawrence joseph henderson and karl albert hasselbalch.

Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.

In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.

How Sciences Interconnect

Medicine: the buffer system in blood.

The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:

The concentration of carbonic acid, H 2 CO 3 is approximately 0.0012 M , and the concentration of the hydrogen carbonate ion, HCO 3 − , HCO 3 − , is around 0.024 M . Using the Henderson-Hasselbalch equation and the p K a of carbonic acid at body temperature, we can calculate the pH of blood:

The fact that the H 2 CO 3 concentration is significantly lower than that of the HCO 3 − HCO 3 − ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.

Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the HCO 3 − HCO 3 − ion, producing H 2 CO 3 . An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO 2 from the blood through the lungs driving the equilibrium reaction such that [H 3 O + ] is lowered. If the blood is too alkaline, a lower breath rate increases CO 2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H + ] and restoring an appropriate pH.

Link to Learning

View information on the buffer system encountered in natural waters.

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Buffer Solution

What is a buffer solution [1-3,5-6], types of buffer solution [6], how to make a buffer solution [2], how to calculate ph of buffer solution [2-4], mechanism of buffer solution [5-6], uses and applications of buffer solution in daily life [6], solved problems.

An aqueous solution that resists any change in pH by adding a small amount of acid or base is called a buffer solution. A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid.

A buffer solution can resist pH change because of an equilibrium between the acid (HA) and its conjugate base (A – ). The balanced equation for this reaction is:

HA ⇋ H + + A –

When a few drops of a strong acid are added to an equilibrium mixture of the weak acid and its conjugate base, the hydrogen ion (H + ) concentration increase very little compared to the volume expected for the quantity of strong acid added. It happens because the equilibrium shifts to the left, following Le Chatelier’s principle. If a strong base is added to the equilibrium mixture, the reaction moves to the right to compensate for the lost H + .

As the buffer’s pH changes slightly when a small amount of strong acid or base is added, it is used to prevent any change in the pH of a solution, regardless of solute. Using buffer solutions in various chemical applications keeps the pH at a nearly constant value.

Characteristics of a Buffer Solution

Some of the characteristics of a buffer solution are as follows:

• It has a definite pH
• Its pH remains the same on standing for a long time
• Its pH does not change on dilution
• Its pH changes negligibly by the addition of a small amount of acids or base

Primarily buffer solutions are of two types: acidic and basic buffers.

Acidic Buffers

A buffer solution prepared with large quantities of a weak acid, and its salt with a strong base, is known as an acid buffer. As the name suggests, these buffer solutions have acidic pH and are used to maintain acidic environments.

A typical example of an acidic buffer is a mixture of an equal concentration of acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa) [pH = 4.74].

Basic Buffers

A buffer solution containing relatively large quantities of a weak base and its salt with a strong acid is called a basic buffer. It has a basic pH and is used to maintain basic conditions.

A typical example of a basic buffer solution would be a mixture of ammonium hydroxide (NH 4 OH) and ammonium chloride (NH 4 Cl) [pH = 9.25].

There are a few ways to prepare a buffer solution with a specific pH.

In the first approach, prepare a solution with acid and its conjugate base by dissolving the acid in about 60% water required to obtain the final volume of the solution. Next, measure the pH of the solution using a pH probe. Then, using a strong base like NaOH, the pH can be adjusted to the desired value. Alternatively, suppose the buffer is prepared with a base and its conjugate acid. In that case, the pH can be altered using a strong acid like HCl. Once the proper pH is achieved, dilute the solution to the final desired volume.

Alternatively, solutions of both the acid form and base form of the solution can be prepared. Both solutions must have the same buffer concentration as the final buffer solution. To achieve the final buffer, add one solution to the other while monitoring the pH.

In the third method, the exact amount of acid and conjugate base needed to make a buffer of a certain pH can be determined using the Henderson-Hasselbach equation: pH = pK a + log[A − ][HA].

To calculate the pH of an acidic or basic buffer solution Henderson-Hasselbalch equation is used.

A weak acid dissociates as follows:

HA(acid) + H 2 O(water)  ⇄  A – (conjugate base) + H 3 O + (hydronium ion)

The Henderson-Hasselbalch equation is given by:

pH = pK a + log 10 [A – ]/[HA]

pH: pH value

K a : acid dissociation constant

[A – ]: Concentration of the conjugate base in moles per liter

[HA]: Concentration of the acid in moles per liter

Similarly, a weak base dissociates as follows:

B + H 2 O  ⇄  OH – + HB +

The Henderson-Hasselbalch equation for a base is give by:

pOH = pK b + log 10 ([HB + ]/[B])

pOH: pOH value

K b : base dissociation constant

[HB + ]: Concentration of the conjugate acid in moles per liter

[B]: Concentration of the base in moles per liter

Factors Determining pH of Buffer Solution

The two factors that determine the pH of the buffer solution are as follows:      

• The equilibrium constant (K a ) of the weak acid: The equilibrium constant of a weak acid reveals the proportion of HA that will be dissociated into H + and A – in solution. If more H + ions are created, the resulting solution becomes more acidic, thus lowering its pH.
• The ratio of weak acid [HA] to weak base [A – ] in solution: If a buffer solution contains more base than acid, more OH – ions are likely to be present, thus raising the pH. On the contrary, if the solution has more acid than base, more H + ions will be present, thus lowering the pH. When the concentrations of HA  and A – are equal, the concentration of H + becomes equal to K a (or equivalently pH = pK a ).

To understand the mechanism of buffer action, let us take an example of an acidic buffer solution of acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa).

In the aqueous medium, they dissociate as:

CH 3 COONa (aq)  ⇌ CH 3 COO – (aq) + Na + (aq) [Completely ionised]

CH 3 COOH (aq)  ⇌ CH 3 COO – (aq) + H + (aq) [Partially ionised]

When a small quantity of strong acid, like HCl, is added to the buffer solution, the additional hydrogen ions combine with the conjugate base (CH 3 COO – ) as follows:

H + (aq) + CH 3 COO – (aq) ↔ CH 3 COOH (aq)

Since the extra H + ions get consumed, the pH of the resulting solution remains constant. As CH 3 COOH is a weak acid and its ions have a strong tendency to form non-ionized CH 3 COOH molecules, the reaction nearly goes to completion.

On the contrary, if a strong base  like NaOH is added, the additional OH – ions get neutralized:

CH 3 COOH (aq) + OH – (aq) → CH 3 COO – (aq) + H 2 O (l)

Also, the added OH – ion reacts with the H + ion to produce water. As a result, the added OH – ions get removed, and the acid equilibrium shifts to the right to replace the used up H + ions. Thus, the pH changes negligibly.

Buffer solutions have various uses and applications in everyday life. Some of them are as follows:

• Maintenance of Life: As most of the biochemical processes in our bodies work within a relatively narrow pH range, the body uses different buffers, such as carbonate and bicarbonate buffer, to maintain a constant pH close to 7.4.
• Biochemical Assays: As enzymes act on a particular pH, buffer solutions keep the pH value constant during an enzyme assay.
• Shampoos : Buffers such as citric acid and sodium hydroxide are used in shampoos to balance out the alkalinity that would otherwise burn our scalp. These buffers counteract the alkalinity of the detergents present in the shampoo.
• Baby Lotions : Baby lotions are buffered to a pH of about 6 to hinder the growth of bacteria within the diaper and also help prevent diaper rash.
• Brewing Industry: Buffer solutions are added before fermentation begins to make sure the solution is not too acidic and prevents the product’s spoiling.
• Textile Industry: Many dyeing processes use buffers to maintain the correct pH for various dyes.
• Laundry Detergents: Many laundry detergents use buffers to prevent their natural ingredients from breaking down.
• Contact Lens : Buffer solutions are used in contact lens solutions to ensure the pH level of the fluid remains compatible with that of the eye.

P.1. Calculate the pH of the solution containing 0.50M sodium acetate and 0.06M acetic acid. (pKa for CH 3 COOH = 5.47)

Soln. According to Henderson-Hasselbalch equation, the pH of the acidic buffer is:

pH=pK a + log[Salt]/[Acid]

Putting the values,

=5.47+log(0.50/0.06)

=5.47+[log0.50−log0.06]

=5.47+(-0.9208)

=5.47-0.9208

P.2. How many moles of sodium acetate and acetic acid each should be dissolved to prepare one liter of 0.063 molar buffer solution of pH 4.5?

(K a ​ for CH 3 ​COOH = 1.8×10 −5 )

Soln. Applying Henderson-Hasselbalch equation

pH=log[Acid][Salt]−logK a

log[Acid][Salt]=4.5+log1.8×10−5=−0.2447

[Salt]=0.5692×[Acid]

[Salt][Acid]=0.063

[Acid]=1.56920.063=0.040M

no of moles =n=0.04×1=0.04mol

[Salt]=(0.063−0.040)=0.023M

∴ no of moles =n=0.023×1=0.023

Ans. As all the reactions between the metal ions and EDTA are pH-dependent, a buffer solution is used in EDTA titration as it resists the change in pH.

Ans. The three major buffer systems of the human body are the carbonic acid bicarbonate buffer system, phosphate buffer system, and protein buffer system.

Ans. To determine the hardness of water, an indicator and EDTA are used. In order to analyze the water, the sample must be kept at a constant pH. As both EDTA and the indicator are weak acids, a buffer solution is used to maintain a reasonably constant pH even when acids and bases are added.

• Introduction to Buffers – Chem.libretexts.org
• Buffer Solutions – Courses.lumenlearning.com
• Buffer Solutions – Chem.purdue.edu
• How can we predict the pH of a buffer? – Chemcollective.org
• Mechanism of buffer action – Qsstudy.com
• Buffer solutions – Researchgate.net

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8.8 Properties of Buffers

4 min read • january 9, 2023

Dylan Black

Jillian Holbrook

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In this section, we review what buffers are and how they form, including why we care about buffers in the first place. For chemistry, we often look for solutions with unique properties, whether that be a specific compound, reaction, or observation made. In the case of buffers , these solutions resist changes in pH. This means that adding strong acids or strong bases to them does not impact the pH as much.

It is important to note, however, that buffers are not immune to changes in pH and do have a certain buffer capacity that we will talk about later. 

Get it? Buffering? We're hilarious. Image from GIPHY

Buffers Review

As we mentioned, buffers are special solutions that are resistant to pH changes when adding acids or bases to them. Buffers are formed in a very specific way: creating a solution of a weak acid and its conjugate base (or a weak base and its conjugate acid , but the former is much more common).

It is important that the acid you create a buffer with is weak because otherwise, the conjugate base would not be a significant base . For example, a mixture of HCl and NaCl would not be a buffer despite being a combination of an acid (HCl) and its conjugate base (Cl-).

You may be asking then why any weak acid isn't a buffer. At equilibrium , there is so much more acid than the conjugate base (assuming a low Ka) that the buffer effects are negligible. In order for a buffer to be effective, you must have comparable concentrations of acid and conjugate base . In fact, the maximum buffer , the point at which the buffer most effectively resists pH change, occurs when the concentration of acid is equal to the concentration of the conjugate base .

Solidify this concept by doing a few practice problems. For each of the pairs of compounds given, identify them as a pair that would form a buffer or not form a buffer:

NaOH and Na+:

The answer to this question is no . Although NaOH and Na+ are a base- conjugate acid pair, remember that NaOH is a strong base . This means that Na+ is not a significant acid and will not form a buffer.

CH3COOH and Ca(CH₃COO)₂:

The answer to this question is yes ! When dissolved together, this pair will form a buffer. CH3COOH is a weak acid (acetic acid AKA vinegar) with a Ka=1.8 * 10^(-5). Ca(CH3COO)2 is calcium acetate, which will dissociate into Ca2+ (a spectator ion as far as the buffer is concerned), and two moles of CH3COO-, the conjugate base of CH3COOH! Because acetic acid is a weak acid , CH3COO- is a significant base , meaning that we will have a buffer.

NH3 and NH4NO3:

This pair does form a buffer. NH3 is a weak base , and NH4+ is a significant acid (and its conjugate acid ), meaning this pair forms a buffer. In this case, like Ca2+ in the previous example, the nitrate ion is simply a spectator.

Like example one, this pair does not form a buffer. HI is a strong acid and cannot form buffers with its conjugate base I- because I- is not a significant base .

KI and PbNO3:

It should be pretty easy to see that this pair does not form a buffer. There are no acids or bases involved. In fact, when you mix KI and PbNO3, you get the "golden rain" reaction, a precipitation reaction that forms PbI2 and KNO3. Take a look!

Image From ChemTalk

What Makes Buffers Cool: pH Resistance

Why do buffers have pH resistance , and what makes them so interesting and useful to study? Buffers have pH resistance because of the presence of an acid and a base that do not actively react together at equilibrium . This graphic shows what happens when an acid or a base is added to a buffer:

Image From LibreTexts

When a strong acid is added to a buffer, the conjugate base eats it up and forms HAn (An = anion ). In the case of no buffer, the strong acid would completely dissociate into H+, increasing [H+] to a much higher degree. Similarly, if OH- from a strong base is added to a buffer, the HAn present in the solution reacts with it to form An- and H2O instead of letting it produce pure OH-. These two reactions lead to buffers being resistant to pH!

Key Terms to Review ( 16 )

Acid Concentration

Base Concentration

Buffer Capacity

Conjugate Acid

Conjugate Base

Equilibrium

pH Resistance

Significant Base

Spectator Ion

Strong Acid

Strong Base

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Module 14: Acid-Based Equilibria

Learning outcomes.

• Describe the composition and function of acid–base buffers
• Calculate the pH of a buffer before and after the addition of added acid or base

A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a  buffer . Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure 1). A solution of acetic acid and sodium acetate (CH 3 COOH + CH 3 COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH 3 ( aq ) + NH 4 Cl( aq )).

Figure 1. (a) The unbuffered solution on the left and the buffered solution on the right have the same pH (pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH. (b) After the addition of 1 mL of a 0.01-M HCl solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)

How Buffers Work

To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, adding strong base to this solution will neutralize hydronium ion and shift the acetic acid ionization equilibrium to the right, partially restoring the decreased H 3 O +  concentration:

[latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\leftrightharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]

Likewise, adding strong acid to this buffer solution will neutralize acetate ion, shifting the above ionization equilibrium right and returning [H 3 O + ] to near its original value. Figure 2 provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer’s conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.

Figure 2. Buffering action in a mixture of acetic acid and acetate salt.

Example 1: pH Changes in Buffered and Unbuffered Solutions

Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.

(a) Calculate the pH of an acetate buffer that is a mixture with 0.10  M  acetic acid and 0.10  M  sodium acetate.

(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.

(c) For comparison, calculate the pH after 1.0 mL of 0.10  M  NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.

To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):

Step 1. Determine the direction of change.

[latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]

We look it up in Ionization Constants of Weak Acids : K a = 1.8 [latex]\times [/latex] 10 −5 .

With [CH 3 CO 2 H] = [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] = 0.10 M and [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = ~0 M , the reaction shifts to the right to form [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex].

Step 3. Solve for x and the equilibrium concentrations.

[latex]x=1.8\times {10}^{-5}M[/latex]

[latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=0+x=1.8\times {10}^{-5}M[/latex]

[latex]\text{pH}=-\text{log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=-\text{log}\left(1.8\times {10}^{-5}\right)[/latex]

[latex]=4.74[/latex]

Step 4. Check the work . If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = K a .

Step 1. Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains

[latex]0.0010\cancel{\text{L}}\times \left(\dfrac{0.10\text{mol NaOH}}{1\cancel{\text{L}}}\right)=1.0\times {10}^{-4}\text{mol NaOH}[/latex]

Step 2. Determine the moles of CH 2 CO 2 H. Before reaction, 0.100 L of the buffer solution contains

[latex]0.100\cancel{\text{L}}\times \left(\dfrac{0.100\text{mol}{\text{ CH}}_{3}{\text{CO}}_{2}\text{H}}{1\cancel{\text{L}}}\right)=1.00\times {10}^{-2}\text{mol}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}[/latex]

Step 3. Solve for the amount of NaCH 3 CO 2 produced. The 1.0 [latex]\times [/latex] 10 −4 mol of NaOH neutralizes 1.0 [latex]\times [/latex] 10 −4 mol of CH 3 CO 2 H, leaving

[latex]\left(1.0\times {10}^{-2}\right)-\left(0.01\times {10}^{-2}\right)=0.99\times {10}^{-2}\text{mol}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}[/latex]

and producing 1.0 [latex]\times [/latex] 10 −4 mol of NaCH 3 CO 2 . This makes a total of

[latex]\left(1.0\times {10}^{-2}\right)+\left(0.01\times {10}^{-2}\right)=1.01\times {10}^{-2}\text{mol}{\text{NaCH}}_{3}{\text{CO}}_{2}[/latex]

Step 4. Find the molarity of the products. After reaction, CH 3 CO 2 H and NaCH 3 CO 2 are contained in 101 mL of the intermediate solution, so:

[latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]=\dfrac{9.9\times {10}^{-3}\text{mol}}{0.101\text{L}}=0.098M[/latex]

[latex]\left[{\text{NaCH}}_{3}{\text{CO}}_{2}\right]=\dfrac{1.01\times {10}^{-2}\text{mol}}{0.101\text{L}}=0.100M[/latex]

This 1.8 [latex]\times [/latex] 10 −5 – M solution of HCl has the same hydronium ion concentration as the 0.10- M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:

[latex]0.100\text{L}\times \left(\dfrac{1.8\times {10}^{-5}\text{mol HCl}}{1\text{L}}\right)=1.8\times {10}^{-6}\text{mol HCl}[/latex]

As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 [latex]\times [/latex] 10 −4 mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:

[latex]\left(1.0\times {10}^{-4}\right)-\left(1.8\times {10}^{-6}\right)=9.8\times {10}^{-5}M[/latex]

• The concentration of NaOH is: [latex]\dfrac{9.8\times {10}^{-5}M\text{NaOH}}{0.101\text{L}}=9.7\times {10}^{-4}M[/latex]
• The pOH of this solution is: [latex]\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]=-\text{log}\left(9.7\times {10}^{-4}\right)=3.01[/latex]
• The pH is: [latex]\text{pH}=14.00-\text{pOH}=10.99[/latex]

The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).

Check Your Learning

Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 [latex]\times [/latex] 10 −5 M HCl solution from 4.74 to 3.00.

Initial pH of 1.8 [latex]\times [/latex] 10 −5 M HCl; pH = −log [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = −log[1.8 [latex]\times [/latex] 10 −5 ] = 4.74

Moles of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] in 100 mL 1.8 [latex]\times [/latex] 10 −5 M HCl; 1.8 [latex]\times [/latex] 10 −5 moles/L [latex]\times [/latex] 0.100 L = 1.8 [latex]\times [/latex] 10 −6

Moles of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L [latex]\times [/latex] 0.0010 L = 1.0 [latex]\times [/latex] 10 −4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:

[latex]\text{pH}=-\text{log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=-\text{log}\left(\dfrac{\text{total moles}{\text{ H}}_{3}{\text{O}}^{\text{+}}}{\text{total volume}}\right)=-\text{log}\left(\dfrac{1.0\times {10}^{-4}\text{mol}+1.8\times {10}^{-6}\text{mol}}{101\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)}\right)=3.00[/latex]

If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 3). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.

Figure 3. The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)

The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.

Selection of Suitable Buffer Mixtures

There are two useful rules of thumb for selecting buffer mixtures:

Figure 4. The graph, an illustration of buffering action, shows change of pH as an increasing amount of a 0.10- M NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH 3 CO 2 H] = 0.10 M and [CH 3 CO 2 − ]=0.10M. Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.

• Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.

Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H 2 CO 3 , and the bicarbonate ion, [latex]{\text{HCO}}_{3}{}^{-}[/latex]. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:

[latex]{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{-}\left(aq\right)\longrightarrow {\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)[/latex]

When an excess of the hydroxide ion is present, it is removed by the reaction:

[latex]{\text{OH}}^{-}\left(aq\right)+{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)\longrightarrow {\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)[/latex]

The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H 3 O +  is converted to H 2 CO 3  and OH –  is converted to HCO 3 – ). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.

The Henderson-Hasselbalch Equation

The ionization-constant expression for a solution of a weak acid can be written as:

[latex]{K}_{\text{a}}=\dfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{-}\right]}{\text{[HA]}}[/latex]

Rearranging to solve for [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex], we get:

[latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]={K}_{\text{a}}\times \dfrac{\text{[HA]}}{\left[{\text{A}}^{-}\right]}[/latex]

Taking the negative logarithm of both sides of this equation, we arrive at:

[latex]-\text{log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=-\text{log}{K}_{\text{a}}\text{- log}\dfrac{\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}[/latex],

which can be written as

• [latex]\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\dfrac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}[/latex]

where p K a is the negative of the common logarithm of the ionization constant of the weak acid (p K a = −log K a ). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation , to calculate the pH of buffer solutions. It is important to note that the “ x is small” assumption must be valid to use this equation.

Lawrence Joseph Henderson and Karl Albert Hasselbalch

Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.

In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.

Medicine: The Buffer System in Blood

The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:

[latex]{\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)\rightleftharpoons {\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)[/latex]

The concentration of carbonic acid, H 2 CO 3 is approximately 0.0012 M , and the concentration of the hydrogen carbonate ion, [latex]{\text{HCO}}_{3}{}^{-}[/latex], is around 0.024 M . Using the Henderson-Hasselbalch equation and the p K a of carbonic acid at body temperature, we can calculate the pH of blood:

[latex]\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\dfrac{\left[\text{base}\right]}{\left[\text{acid}\right]}=6.1+\text{log}\dfrac{0.024}{0.0012}=7.4[/latex]

The fact that the H 2 CO 3 concentration is significantly lower than that of the [latex]{\text{HCO}}_{3}{}^{-}[/latex] ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.

Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the [latex]{\text{HCO}}_{3}{}^{-}[/latex] ion, producing H 2 CO 3 . An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO 2 from the blood through the lungs driving the equilibrium reaction such that [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] is lowered. If the blood is too alkaline, a lower breath rate increases CO 2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H + ] and restoring an appropriate pH.

Key Concepts and Summary

Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.

Key Equations

• [latex]\text{p}K_{\text{a}}=−\text{log}K_{\text{a}}[/latex]
• [latex]\text{p}K_{\text{b}}=−\text{log}K_{\text{b}}[/latex]
• Explain why a buffer can be prepared from a mixture of NH 4 Cl and NaOH but not from NH 3 and NaOH.
• Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H 3 PO 4 and a salt of its conjugate base NaH 2 PO 4 .
• Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH 3 and a salt of its conjugate acid NH 4 Cl.
• What is [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] in a solution of 0.25 M CH 3 CO 2 H and 0.030 M NaCH 3 CO 2 ? [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right){K}_{\text{a}}=1.8\times {10}^{-5}[/latex]
• What is [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] in a solution of 0.075 M HNO 2 and 0.030 M NaNO 2 ? [latex]{\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NO}}_{2}{}^{-}\left(aq\right){K}_{\text{a}}=4.5\times {10}^{-5}[/latex]
• What is [OH − ] in a solution of 0.125 M CH 3 NH 2 and 0.130 M CH 3 NH 3 Cl? [latex]{\text{CH}}_{3}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=4.4\times {10}^{-4}[/latex]
• What is [OH − ] in a solution of 1.25 M NH 3 and 0.78 M NH 4 NO 3 ? [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=1.8\times {10}^{-5}[/latex]
• What concentration of NH 4 NO 3 is required to make [OH − ] = 1.0 [latex]\times [/latex] 10 −5 in a 0.200- M solution of NH 3 ?
• What concentration of NaF is required to make [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 2.3 [latex]\times [/latex] 10 −4 in a 0.300- M solution of HF?
• CH 3 CO 2 H
• What will be the pH of a buffer solution prepared from 0.20 mol NH 3 , 0.40 mol NH 4 NO 3 , and just enough water to give 1.00 L of solution?
• Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH 2 PO 4 , and enough water to make 0.500 L of solution.
• How much solid NaCH 3 CO 2 •3H 2 O must be added to 0.300 L of a 0.50- M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
• What mass of NH 4 Cl must be added to 0.750 L of a 0.100- M solution of NH 3 to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)

2. Excess [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] is removed primarily by the reaction [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{H}}_{2}{\text{PO}}_{4}{}^{-}\left(aq\right)\longrightarrow {\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)[/latex]

Excess base is removed by the reaction [latex]{\text{OH}}^{-}\left(aq\right)+{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)\longrightarrow {\text{H}}_{2}{\text{PO}}_{4}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)[/latex]

4. The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=1.8\times {10}^{-5}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.25 − x ) ≈ 0.25 and (0.030 − x ) ≈ 0.030, gives:

[latex]\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\frac{\left(x\right)\left(0.030-x\right)}{\left(0.25-x\right)}\approx \frac{\left(x\right)\left(0.030\right)}{0.25}=1.8\times {10}^{-5}[/latex]

Solving for x gives 1.50 [latex]\times [/latex] 10 −4 M . Because this value is less than 5% of both 0.25 and 0.030, our assumptions are correct. Therefore, [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 1.5 [latex]\times [/latex] 10 −4 M .

This problem can also be solved using the Henderson-Hasselbalch equation: [latex]\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}[/latex]; p K a = −log( K a ) = −log(1.8 [latex]\times [/latex] 10 −5 ) = 4.74; [HA] ≈ [HA] 0 = [CH 3 CO 2 H] 0 = 0.25 M ; [A − ] ≈ [NaCH 3 CO 2 ] = 0.030 M . Using these data: [latex]\text{pH}=4.74-\text{log}\left(\frac{0.030M}{0.25M}\right)=3.82[/latex]; [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 10 −pH M = 10 −3.82 M = 1.5 [latex]\times [/latex] 10 −4 M

6. The equilibrium expression is: [latex]{K}_{\text{b}}=\frac{\left[{\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{3}{\text{NH}}_{2}\right]}=4.4\times {10}^{-4}[/latex]

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.125 − x ) ≈ 0.125 and (0.130 − x ) ≈ 0.130, gives:

[latex]\frac{\left[{\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{3}{\text{NH}}_{2}\right]}=\frac{\left(0.130-x\right)\left(x\right)}{\left(0.125-x\right)}\approx \frac{\left(0.130\right)\left(x\right)}{0.125}=4.4\times {10}^{-4}[/latex]

Solving for x gives 4.23 [latex]\times [/latex] 10 −4 M . Because this value is less than 5% of both 0.125 and 0.130, our assumptions are correct. Therefore, [OH − ] = 4.2 [latex]\times [/latex] 10 −4 M .

8. The reaction and equilibrium constant are [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,{K}_{\text{b}}=1.8\times {10}^{-4}[/latex]

The equilibrium expression is [latex]{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=1.8\times {10}^{-5}[/latex]

Let x = the concentration of NH 4 NO 3 required. The initial and equilibrium concentrations for this system can be written as follows:

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that ( x + x ) ≈ x , gives:

[latex]\frac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(x-x\right)\left(1.0\times {10}^{-5}\right)}{\left(0.200 - 1.0\times {10}^{-5}\right)}\approx \frac{\left(x\right)\left(1.0\times {10}^{-5}\right)}{0.200}=1.8\times {10}^{-5}[/latex]

Solving for x gives 0.360 M . Because x is less than 5% of this value, our assumption is correct. Therefore, [latex]\left[{\text{NH}}_{4}{}^{\text{+}}\right][/latex] = [NH 4 NO 3 ] = 0.36 M .

10. The reaction and equilibrium constant are [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,{K}_{\text{a}}=1.8\times {10}^{-5}[/latex]

• (a) The added HCl will increase the concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] slightly, which will react with [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex] and produce CH 3 CO 2 H in the process. Thus, [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] decreases and [CH 3 CO 2 H] increases.
• (b) The added KCH 3 CO 2 will increase the concentration of [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] which will react with [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and produce CH 3 CO 2 H in the process. Thus, [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] decreases slightly and [CH 3 CO 2 H] increases.
• (c) The added NaCl will have no effect on the concentration of the ions.
• (d) The added KOH will produce OH − ions, which will react with the [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex], thus reducing [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex]. Some additional CH 3 CO 2 H will dissociate, producing [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] ions in the process. Thus, [CH 3 CO 2 H] decreases slightly and [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] increases.
• (e) The added CH 3 CO 2 H will increase its concentration, causing more of it to dissociate and producing more [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] and [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] in the process. Thus, [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] increases slightly and [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] increases.

12. The reaction and equilibrium constant are: [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,{K}_{\text{b}}=1.8\times {10}^{-5}[/latex]

The equilibrium expression is: [latex]{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=1.8\times {10}^{-5}[/latex]

The initial concentrations of NH 3 and [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] are 0.20 M and 0.40 M , respectively. The equilibrium concentrations for this system can be written as follows:

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.20 − x ) ≈ 0.20 and (0.40 + x ) ≈ 0.40, gives:

[latex]\frac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(0.40+x\right)\left(x\right)}{\left(0.20-x\right)}\approx \frac{\left(0.40\right)\left(x\right)}{0.20}=1.8\times {10}^{-5}[/latex]

Solving for x gives 9.00 [latex]\times [/latex] 10 −6 M . Because this value is less than 5% of both 0.20 and 0.40, our assumptions are correct. Therefore, [OH − ] = 9.00 [latex]\times [/latex] 10 −6 M . Thus:

• pOH = −log(9.00 [latex]\times [/latex] 10 −6 ) = 5.046
• pH = 14.000 − pOH = 14.000 − 5.046 = 8.954 = 8.95

14. The reaction and equilibrium constant are: [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,{K}_{\text{a}}=1.8\times {10}^{-5}[/latex]

The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=1.8\times {10}^{-5}[/latex]

Let x be the concentration of [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex]. The hydronium ion concentration at equilibrium is [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 10 −pH = 10 −5.00 = 1.00 [latex]\times [/latex] 10 −5 M

[latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\frac{\left(1.0\times {10}^{-5}\right)\left(x+x\right)}{\left(0.50 - 1.0\times {10}^{-5}\right)}\approx \frac{\left(1.0\times {10}^{-5}\right)\left(x\right)}{0.50}=1.8\times {10}^{-5}[/latex]

Solving for x gives 0.900 M . Because x is less than 5% of this value, our assumption is correct. Therefore, [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right][/latex] = 0.900 M . Using the molar mass of NaC 2 H 3 O 2 •3H 2 O (136.080 /mol) and the volume gives the mass required:

[latex]\frac{0.900\text{mol}}{1\text{L}}\times 0.300\text{L}\times \frac{136.080\text{g}}{1\text{mol}}=36.7=37\text{g}\left(0.27\text{mol}\right)[/latex]

• What is the pH of the solution?
• Is the solution acidic or basic?
• What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
• What is the pH of this buffer solution?
• What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?

The reaction and equilibrium constant are [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,{K}_{\text{a}}=1.8\times {10}^{-5}[/latex]

The molar mass of NH 4 Cl is 53.4912 g/mol. The moles of NH 4 Cl are: [latex]\frac{5.36\text{g}}{53.4912\text{g}{\text{mol}}^{-1}}=0.1002\text{mol}[/latex]

Assume 0.500 L of each solution is present The total volume is thus 1.000 L. The initial concentrations of the ions is obtained using M 1 V 1 = M 2 V 2 , or:

[latex]\begin{array}{l} \left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]={M}_{1}\times \frac{{V}_{1}}{{V}_{2}}=\left(0.200\right)\times \frac{0.500\text{L}}{1.000\text{L}}=0.100M\\ \left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]={M}_{1}\times \frac{{V}_{1}}{{V}_{2}}=\left(0.600\right)\times \frac{0.500\text{L}}{1.000\text{L}}=0.300M\end{array}[/latex]

The initial and equilibrium concentrations of this system can be written as follows:

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.100 − x ) ≈ 0.100 and (0.300 − x ) ≈ 0.300, gives:

[latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\frac{\left(x\right)\left(0.300+x\right)}{\left(0.100-x\right)}\approx \frac{\left(x\right)\left(0.300\right)}{0.100}=1.80\times {10}^{-5}[/latex]

Solving for x gives 6.000 [latex]\times [/latex] 10 −6 M . Because this value is less than 5% of both 0.100 and 0.300, our assumptions are correct. Therefore [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 6.000 [latex]\times [/latex] 10 −6 M :

pH = −log(6.000 [latex]\times [/latex] 10 −6 ) = 5.2218 = 5.222;

The solution is acidic.

Assume that the added H + reacts completely with an equal amount of [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex], forming an equal amount of CH 3 CO 2 H in the process. The moles of H + added equal 0.034 M [latex]\times [/latex] 0.00300 L = 1.02 [latex]\times [/latex] 10 −4 mol. For the acetic acid, the initial moles present equal 0.2000 M [latex]\times [/latex] 0.500 L = 0.1000 mol, and for acetate ion, 0.600 M [latex]\times [/latex] 0.500 L = 0.3000 mol. Thus:

mol CH 3 CO 2 H = 0.1000 + 1.02 [latex]\times [/latex] 10 −4 = 0.1001 mol

[latex]\text{mol}{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}=0.3000 - 1.02\times {10}^{-4}=0.2999\text{mol}[/latex]

Final volume = 1.000 L + 3.00 [latex]\times [/latex] 10 −3 L = 1.0030 L

The initial concentrations are therefore:

• [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]=\frac{0.1001\text{mol}}{1.0030\text{L}}=0.09980M[/latex]
• [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]=\frac{0.2999\text{mol}}{1.0030\text{L}}=0.2990M[/latex]

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.09980 − x ) ≈ 0.09980 and (0.2990 − x ) ≈ 0.2990, gives:

[latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\frac{\left(x\right)\left(0.2990+x\right)}{\left(0.09980-x\right)}\approx \frac{\left(x\right)\left(0.2990\right)}{0.09980}=1.80\times {10}^{-5}[/latex]

Solving for x gives 6.008 [latex]\times [/latex] 10 −6 M . Because this value is less than 5% of both 0.09980 and 0.2990, our assumptions are correct. Therefore, [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 6.008 [latex]\times [/latex] 10 −6 M .

pH = −log(6.008 [latex]\times [/latex] 10 −6 ) = 5.2213 = 5.221

• Which acid in Table 1 of  Relative Strengths of Acids and Bases  is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.
• Which acid in Table 1 of  Relative Strengths of Acids and Bases  is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.
• Which base in Table 2 of  Relative Strengths of Acids and Bases  is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.
• Which base in Table 2 of  Relative Strengths of Acids and Bases is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.
• Saccharin, C 7 H 4 NSO 3 H, is a weak acid ( K a = 2.1 [latex]\times [/latex] 10 −2 ). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 [latex]\times [/latex] 10 −3 g of sodium saccharide, Na(C 7 H 4 NSO 3 ), what are the final concentrations of saccharine and sodium saccharide in the solution?
• What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C 5 H 9 NO 4 , a diprotic acid; K 1 = 8.5 [latex]\times [/latex] 10 −5 , K 2 = 3.39 [latex]\times [/latex] 10 −10 ) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?

1. To prepare the best buffer for a weak acid HA and its salt, the ratio [latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{{K}_{\text{a}}}[/latex] should be as close to 1 as possible for effective buffer action. The [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] concentration in a buffer of pH 3.1 is [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 10 −3.1 = 7.94 [latex]\times [/latex] 10 −4 M

We can now solve for K a of the best acid as follows:

[latex]\begin{array}{l}{ }\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{{K}_{\text{a}}}=1\\ {K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{1}=7.94\times {10}^{-4}\end{array}[/latex]

In Table 1 of  Relative Strengths of Acids and Bases , the acid with the closest K a to 7.94 [latex]\times [/latex] 10 −4 is HF, with a K a of 7.2 [latex]\times [/latex] 10 −4 .

3. For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio [latex]\frac{\left[{\text{OH}}^{-}\right]}{{K}_{\text{b}}}[/latex] that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH − ] is [OH − ] = 10 −pOH = 10 −3.35 = 4.467 [latex]\times [/latex] 10 −4 M .

We can now solve for K b of the best base as follows:

[latex]\frac{\left[{\text{OH}}^{-}\right]}{{K}_{\text{b}}}=1[/latex]

K b = [OH − ] = 4.47 [latex]\times [/latex] 10 −4

In Table 2 of  Relative Strengths of Acids and Bases , the base with the closest K b to 4.47 [latex]\times [/latex] 10 −4 is CH 3 NH 2 , with a K b = 4.4 [latex]\times [/latex] 10 −4 .

5. The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is: [latex]2.00\times {10}^{-3}\text{g}\times \frac{1\text{mol}}{205.169\text{g}}=9.75\times {10}^{-6}\text{mol}[/latex]

This ionizes initially to form saccharin ions, A − , with:

[latex]\left[{\text{A}}^{-}\right]=\frac{9.75\times {10}^{-6}\text{mol}}{0.250\text{L}}=3.9\times {10}^{-5}M[/latex]

but A − reacts with water:

[latex]\begin{array}{l}{\text{A}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons \text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ {K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{2.1\times {10}^{-12}}=4.8\times {10}^{-3}\\ =4.8\times {10}^{-3}=\frac{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{A}}^{-}\right]}\end{array}[/latex]

The pH of the solution is 5.48, so pOH = 14.00 − 5.48 = 8.52, and [OH − ] = 10 −8.52 = 3.02 [latex]\times [/latex] 10 −9 M

Because of the small size of K b , almost all the A − will be in the form of HA. Therefore, [latex]4.8\times {10}^{-3}=\frac{x\left(3.02\times {10}^{-9}\right)}{3.9\times {10}^{-5}-x}[/latex], where  x ≈ 3.9 [latex]\times [/latex] 10 −5 M = [HA] = [C 7 H 4 NSO 3 H]

Consequently, [A − ] is extremely small. Therefore, solve for [A − ] from the equilibrium expression:

[latex]\left[{\text{A}}^{-}\right]=\frac{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]}{{K}_{\text{b}}}=\frac{\left(3.9\times {10}^{-5}\right)\left(3.02\times {10}^{-9}\right)}{4.8\times {10}^{-3}}=2.5\times {10}^{-11}M=\left[\text{Na}\left({\text{C}}_{7}{\text{H}}_{4}{\text{NSO}}_{3}\right)\right][/latex]

buffer capacity: amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)

buffer: mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added

Henderson-Hasselbalch equation: equation used to calculate the pH of buffer solutions

• Chemistry 2e. Provided by : OpenStax. Located at : https://openstax.org/ . License : CC BY: Attribution . License Terms : Access for free at https://openstax.org/books/chemistry-2e/pages/1-introduction

Buffer Solution

A buffer solution is a chemical solution that resists change to its pH or acidity. It is one that resists changes in pH when small quantities of an acid or an alkali are added to it. It is a solution in water of a mixture of a weak acid or base and its salt. The pH of the solution changes very little when a small amount of strong acid or base is added to it. It only means that the change in pH is not as much as it would be with a solution that is not a buffer.

Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. A lot of biological and chemical reactions need a constant pH for the reaction to proceed. Many life forms have a relatively small pH range; an example of a buffer solution is blood. Buffers are extremely useful in these systems to maintain the pH at a constant value. They are necessary for biology for keeping the correct pH for proteins to work.

Buffer solutions may be of two types: acidic and basic.

• Acidic: A solution of a weak acid and its salt. Acidic buffer solutions are commonly made from a weak acid and one of its salts – often a sodium salt.
• Basic: A solution of a weak base and its salt. An alkaline buffer solution has a pH greater than 7. A frequently used example is a mixture of ammonia solution and ammonium chloride solution.

If you add acid to the solution, the concentration of H+ ions will increase; to keep equilibrium a small number of ions will be combined (forming a salt and reducing the concentration of H+ ion in the solution). The use of one or the other will simply depend upon the desired pH when preparing the buffer. If you add base the concentration of H+ ion will reduce (by consumption or combining) and so a small amount of salt will break into ions and maintain the pH.

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1.7: pH and Buffers

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• Page ID 36749

• Orange County Biotechnology Education Collaborative
• ASCCC Open Educational Resources Initiative

Learning Objectives

• Accurately measure the pH of solutions using pH indicator strips and a pH meter.
• Create buffer solutions and test the effects of adding acid and base to each.

Student Learning Outcomes:

Upon completion of this lab, students will be able to:

• Describe the pH scale.
• Correctly use pH indicator strips and a pH meter.
• Explain the function and composition of a buffer.

Introduction

The pH of solutions is an important characteristic. Cells must maintain a constant pH so that the enzymes and processes taking place inside the cells will continue as needed. Chemical and enzymatic reactions are typically dependent on a specific pH range. Thus, it is important to understand pH and be able to determine the pH of various solutions.

The pH scale is a familiar concept for students who study science. The pH value of a solution reflects the relative concentration of hydrogen ions (H+) or protons to the concentration of hydroxide ions (OH-) in a solution. Solutions with a pH value less than 7 are acidic and those with a value greater than 7 are basic, or alkaline. The value 7 is neutral meaning the amount of H+ in a solution is equal to the amount of OH- in a solution. Pure water H 2 O, which can dissociate naturally into H + and OH - ions, would have a value of 7.

\[\ce{H2O <=> H^{+} + OH^{-}} \nonumber\]

Pre-lab Reading Assignment:

Chemistry review.

In a chemical equation, variables that are surrounded by brackets “[“ and “]” are expressions of concentration, or the specific amount of a molecule in a given volume of solution. For example, if you see “[H+]” in an equation, this is read as “the concentration of hydrogen ion”.

The concentration of a solution is often expressed in units of moles per liter (mol/L). Just as one “dozen” represents a quantity of 12 items, one “mole” represents a quantity of approximately 6.022 X 10 23 items.

one dozen molecules = 12 molecules

one mole of molecules = 602,200,000,000,000,000,000,000 molecules!

Note: “ n ” is used in equations to indicate a quantity measured in moles. For example if you see “ n Acid ” in an equation, this is read as “moles of acid”.

The term “Molarity” indicates that a solution’s concentration is in units of moles per liter. A one molar solution (1 M) contains one mole of solute within each liter of that solution. Reagents used in the laboratory will often be labeled with their concentrations expressed in terms of molarity.

The relative concentration of H+ or OH- may change very dramatically in solutions, so a logarithmic scale (called pH) instead of a linear scale is used to express concentration. Equations 2 and 3 can be used to calculate the pH based on hydrogen ion concentration or vice versa.

To calculate pH based on hydrogen ion concentration [H + ]:

pH = -log [H + ]

To calculate hydrogen ion concentration [H + ] based on pH:

[H + ] = 10 -pH

A buffer is a mixture of a weak acid (HA) and its salt (e.g., NaA), and is sometimes referred to as a conjugate acid-base pair . As mentioned above, buffers have a major role in stabilizing the pH of living systems. Vertebrate organisms maintain the pH of blood using a buffer composed of a mixture of carbonic acid (H 2 CO 3 ) and sodium bicarbonate (Na + HCO 3 - ). The weak acid in this buffer is carbonic acid and the salt is sodium bicarbonate. When dissolved in water, sodium bicarbonate disassociates completely into sodium ions (Na + ) and bicarbonate ions (HCO 3 - ). The H 2 CO 3 is the conjugate acid of HCO 3 - and the HCO 3 - is the conjugate base of H 2 CO 3 . Together, this conjugate acid-base pair functions as the bicarbonate buffer system.

Buffer systems are also of particular importance to experimental cell biology.

The pH of a buffer solution may be calculated as follows:

\[pH=pK_a + log \frac{n_A}{n_{HA}}\nonumber\]

Where pK a = dissociation constant of the acid, n A = initial number of moles of salt in the buffer, and n HA = initial number of moles of acid in the buffer.

If you know these values, it is possible to accurately calculate the pH of a buffer system before you create it!

The pK a of acetic acid (used in today’s experiment) is 4.75

To find the volume of the conjugate base or conjugate acid:

n A = volume of conjugate base (mL) \( \times \dfrac{1\: L}{100\: mL} \times \) concentration of conjugate base (mol/L)

n H A = volume of conjugate acid (mL) \( \times \dfrac{1\: L}{100\: mL} \times \) concentration of conjugate acid (mol/L)

Use of pH Indicator Strips

The pH of a solution can be roughly approximated using strips of paper treated with color changing indicator reagents. The strips are dipped into the solution to be tested for several seconds and then removed. The color of the indicator strip is then compared to a reference chart, often printed on the side of the strip’s container. The reference color on the chart that most closely matches the color of the reacted strip will have a pH value printed below it and that will be the approximate pH. One advantage to using pH indicator strips is that they are relatively inexpensive, easy to use, and are adequate for determining pH where an error of +/- 1 pH unit is acceptable. A more accurate method of determining pH is to use a calibrated pH meter, which can determine the exact pH to one or more decimal places depending on the quality of the device.

Use of a pH Meter

The pH meter measures the acidity of a solution. It is a scientific instrument that uses electrodes to measure the hydrogen ion (proton) concentration of water-based solutions. Essentially, the pH meter is a voltmeter that will measure the difference between two electrodes. The probe you place into the solution contains a reference electrode and a detector electrode. The reference electrode is not affected by the solution being measured and is in contact with a solution of potassium chloride. The detector electrode comes in contact with the test solution. The hydrogen ions in the test solution interact with the electrode and the difference in electrical potential between the two electrodes is detected and reported as millivolts or converted to a pH value.

For accurate measurements, it is important to calibrate your pH meter before use with buffer solutions of known values. It is best to calibrate your meter with buffer solutions that are near the anticipated or desired pH of your test solution. You should also blot the probe with laboratory wipes in between solutions to avoid contamination but avoid rubbing. Rubbing the probe may cause a static electricity charge to build up on the electrode which will cause inaccurate readings to occur. Accidentally letting the probe dry out will also cause it to stop working so always keep the end of the probe immersed in a holding solution when not taking measurements. Remember to return it to the storage solution as soon you are finished with the experiment.

Calibrate the pH meter for pH 4, 7, and 10 before taking measurements. If calibrated properly, your pH meter should produce measurements with an accuracy of +/- 0.06 pH units. Always test your meter after calibration using the standard buffers and recalibrate the meter if necessary before proceeding.

Your instructor will demonstrate the proper calibration, care, and use of the meter. Be sure to take good notes!

Activity 1: Measuring pH

Per group of 4:

• 1 Set of 4 unknown solutions (in 30 mL tubes with screw top lids)
• 1 container of pH indicator strips and color reference chart
• 1 pair of forceps
• 1 pH meter (calibrated – See instructor for directions)
• Obtain a set of unknown solutions from instructor.
• Measure the pH of each solution using the pH indicator strips first. Hold the strips with the forceps. Use a new strip for each solution!
• Record your data in Table 1.
• Measure the pH of each solution using the pH meter. Be sure to rinse the tip of the probe with DI water before putting the probe into each sample! (Ask the instructor for instructions if you are not sure how to properly calibrate and use the pH meter).
• Record your data in Table 2.

Data Analysis

• How do your pH indicator strip values compare to your pH meter values?
• Check your measured pH values with those of the other teams. Are your values similar?
• Check with your instructor to see what the actual pH values should be. How accurate were you?

Activity 2: Preparation of an Acetate Buffer

• 1 bottle stock solution of 0.1 M acetic acid (CH 3 COOH)
• 1 bottle stock solution of 0.1 M sodium acetate (Na + CH 3 COO - )

Per Group of 4:

• 6 clean 30 mL plastic tubes
• 2 clean 5 mL serological pipettes
• 2 pipette pumps (10 mL capacity)
• 1 Sharpie Marker
• Using a sharpie marker, label the two 30 mL tubes - one as “Acetic Acid” and the other “Sodium Acetate”. Fill each tube up with the correct stock solution.
• Using a sharpie marker, label each of the two 5 ml pipettes - one as “AA” and the other as “SA”. To avoid contamination, DO NOT dip pipettes into stock solution bottles and ONLY use the designated pipette to transfer either acetic acid or sodium acetate from your group’s labeled tubes.
• Using a sharpie marker, label a clean 30 mL tube as “Buffer 1”, another as “Buffer 2”, the third as “Buffer 3”, and the fourth as “H2O”. Each student in your group will take one tube. If there are only 3 students, one of you can also take the “H2O” tube. Write your names into the first column of table 2 next to the tube(s) you will be working with.
• Be sure to accurately pipet the volumes indicated to get good results! Review proper pipetting technique with your instructor if necessary.
• For the “H2O” tube, simply fill the tube about a third full with pure deionized water
• Close the lids and gently shake each tube for about 20 seconds or more to mix the contents.
• Measure the pH of each solution with the pH meter using proper technique and enter your measurements in table 2.

Activity 3: Effects of Adding Acid and Base to Acetate Buffer

• Everything From Activity (2 above)
• 30 mL dropper bottle of 0.1 M HCl (Hydrochloric Acid)
• 30 mL dropper bottle of 0.1 M NaOH (Sodium Hydroxide)
• Add a single drop of HCl to each of your team’s 4 tubes. Close the lids and gently shake the tubes to thoroughly mix the contents.
• Measure the pH of each solution and enter the pH values in table 4.
• Continue adding drops of HCl according to the table, measuring pH, and recording values.
• When you have completed Table 3, you will now start adding drops of NaOH (base) to your tubes according to table 5.
• Be sure to shake the tubes to mix the contents thoroughly before measuring pH and entering the values in Table 5.
• Look at your results and compare the pH changes in your 4 tubes. What do you notice about the pH changes when you compare them?
• Compare your pH values to those of the other teams. Ask your instructor for the expected values.

Study Questions

• What range of pH values indicates that a solution is acidic? Basic?
• In general, how does the relative concentration of hydrogen ions [H+] compare to that of hydroxide ions [OH-] in a neutral, acidic, and basic solution?
• Based on your observations, how would you describe what a buffer does?
• What factors determine the accuracy of a reading with a pH meter?
• What is the pH of a solution that has a hydrogen ion concentration of 2.46 X 10 -5 M?
• What is the expected pH of a buffer made from 25.7 mL of 2.0 M Acetic acid and 0.0492 L of 0.90-M Sodium acetate?

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• Buffer Solution

Introduction: What is a Buffer Solution?

A buffer is an aqueous solution that consists of a mixture of a weak acid and its salt (acid buffer) or a weak base with its salt (basic buffer). Its pH changes very little when a small amount of strong acid or base is added to it and is thus used to prevent a solution's pH change.

Buffer solutions are used for a wide range of chemical applications. Blood is one example of a buffer solution found in nature. Human blood has a natural pH of 7.4. Many people experience severe anxiety and suffer from alkalosis. Alkalosis is a disease in which blood pH is excessively high. The reverse condition is called acidosis-a blood, pH greater than 7.4

Some chemical reactions only occur at a certain pH. Other households and consumer items need to monitor their pH values, such as shampoo to combat the soap's alkalinity to avoid inflammation, baby lotion to retain a pH of around 6 to discourage multiplication of bacteria, washing powder, eye drops, fizzy lemonade etc.

Buffer Solution Definition

Solutions with the stable concentration of hydrogen ions and thus typically with no change in pH which is almost independent of dilution and which change very little with small additions of a strong acid or alkali are called buffers. It can also be described in simple terms as a solution that prevents any pH change when a small amount of a strong acid or a strong base is applied to it, which is called a buffer solution or simply as a buffer. Both buffers have acidity and alkalinity balance.

Any compounds, such as ammonium acetate, tend to resist any change in their concentration of hydronium ions or pH, whenever a small amount of a strong acid or a strong base is applied to it.

Buffer solutions usually consist of a mixture of a weak acid and salt with a strong base like CH 3 COOH and CH 3 COONa, or a weak base with a strong acid like NH 4 OH and NH 4 Cl and salt.

Mechanism of Buffering Action

Consider the example of a buffer solution made by dissolving sodium acetate into acetic acid, to consider how a buffer functions. As you can see from the name, acetate acid is an acid: CH 3 COOH, while sodium acetate dissociates in solution to yield the conjugate base, CH 3 COO-acetate ions. The reaction equation is:

CH 3 COOH (aq) + OH-(aq) 🡪CH 3 COO-(aq) + H 2 O (aq)

If this solution is combined with a strong acid, the acetate ion can neutralise.

CH 3 COO-(aq) + H+(aq) 🡪CH 3 COOH (aq)

It changes the original buffer reaction equilibrium, thereby holding the pH steady.

Preparation of Buffer Solution

There are a few methods to prepare a buffer solution with a different pH. Prepare a solution with acid and its conjugate base in the first approach by dissolving the acid component of the buffer in around 60 per cent of the amount of water used to produce the final volume of solution.

Instead, use a pH detector to test the pH of the solution. Using a strong base like NaOH the pH can be changed to the desired value. If a base and its conjugate acid are used to make the buffer, the pH can be modified using a strong acid, like HCl. Dilute the solution to the final desired volume, once the pH is right.

Additionally, you should prepare solutions for both the solution's acid type and base form. Both solutions must have the same quantity of buffer as in the final solution. Add one solution to the other while tracking the pH to get the final buffer.

In a third method, using the Henderson-Hasselbach equation, you can determine the exact amount of acid and conjugate base required to make a buffer of a certain pH:

 pH = pKa + log|A−||HA|

Types of Buffer Solution

There are two buffer forms, acid buffer, and base buffer.

Acid Buffer

A buffer solution that contains large quantities of a weak acid, and its salt with a strong base, is called an acid buffer. On the acidic side, such buffer solutions have pH, i.e.pH is below 7 at 298 K. The equation gives the pH of an acid buffer. CH 3 COOH, with CH 3 COONa.

pH = pKa + ln(Salt)Acid

Where Ka -----acid dissociation constant of the weak acid

Basic Buffer

A buffer solution that contains relatively large quantities of a weak base and its salt with a strong acid is called a simple buffer. On the alkaline side, these buffers have pH, i.e., pH is higher than 7 at 298 K. For example, NH 4 OH and NH 4 Cl.

The pH of an appropriate buffer is determined by the equation

pOH = pKb + ln(Salt)Acid

Where, Kb ------base dissociation constant.

These equations are called Henderson Hasselbalch equations

Buffer Solution Examples

Blood - contains a bicarbonate buffer system

Tris buffer

Phosphate buffer

As mentioned, buffers are beneficial over specific pH ranges. For example, here is the pH range of common buffering agents:

While making a buffer solution, the pH of the solution is changed to get it within the right effective range. A strong acid, such as hydrochloric acid (HCl), is usually added to reduce the pH of acidic buffers. A strong base such as sodium hydroxide (NaOH) solution is added to increase the pH of the alkaline buffers.

Importance of Buffers

The acidity of the solution in which they occur affects a lot of chemical reactions. The pH of the reaction medium must be controlled for a given reaction to occur or to occur at a suitable rate. This control is provided by buffer solutions, which are solutions that preserve a certain pH. Biochemical reactions are particularly sensitive to pH. Most biological molecules contain groups of atoms that can be charged or neutral based on pH, and whether these groups are charged or neutral has a significant effect on the molecule's biological activity.

The fluid within the cell and the fluids around the cells have a characteristic and almost constant pH in all multicellular organisms. This pH is preserved in several ways, and one of the most important is through buffer systems.

FAQs on Buffer Solution

1. What are Buffer Solutions Examples?

Acid buffers are liquids with a pH of below 7, containing a weak acid and one of its salts. A combination of acetic acid and sodium acetate for example serves as a buffer solution with a pH of about 4.75.

2. What are Buffer Solutions Used for?

Buffer solutions are used in a wide variety of chemical applications as a means of keeping pH to an almost constant value. There are many systems in nature which use buffering for regulating pH. The bicarbonate buffering system for example is used to regulate the pH of the blood.

3. How Do You Make a Buffer Solution?

Add water to 1 litre. (Alternatively, dilute 10 times 100 mM of phosphoric acid (sodium) buffer solution (pH=6.8). Add water to 1 litre. Add water to 1 litre.

4. Why are Buffer Solutions Important?

A buffer is a solution that can tolerate pH change when an acidic or basic component is applied. It can neutralise small amounts of added acid or base and thus retain a fairly steady pH of the solution. This is important for processes and/or reactions where unique and stable pH ranges are needed.

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5. What is a buffer solution?

A buffer solution is an aqueous mixture of a weak acid and its conjugate base. When a normal quantity of strong acid or base is introduced to it, the pH hardly changes. Buffer solutions are widely used in chemical applications to keep pH at a constant value. Many natural systems rely on buffering to maintain pH balance. The bicarbonate buffering mechanism, for example, is used to maintain blood pH, and bicarbonate also serves as a buffer in the ocean. Visit Vedantu website to learn more. 

6. What are the applications of buffers?

Regardless of what else is in the solution, the pH of a solution containing a buffering agent can only vary within a narrow range. This is a requirement for enzymes to function properly in biological systems. The plasma component of human blood, for example, contains a mixture of carbonic acid and bicarbonate, which is the fundamental mechanism for keeping blood pH between 7.35 and 7.45. Acidosis and alkalosis metabolic states occur quickly outside of this restricted range (7.40 0.05 pH unit), eventually leading to death if the proper buffering capacity is not quickly restored.

The efficacy of an enzyme declines when the pH of a solution rises or falls too much, a process known as denaturation, which is usually irreversible. The bulk of biological samples used in research is stored in a buffer solution, which is usually phosphate-buffered saline (PBS) with a pH of 7.4.

Buffering agents are used in the industry in fermentation processes and to set the proper conditions for dyes used in fabric colouring. They are also employed in chemical analysis and pH metre calibration.

The pH of buffers in acidic environments can be changed to a desirable value by adding a strong acid to the buffering agent, such as hydrochloric acid. A strong base, such as sodium hydroxide, can be used to make alkaline buffers. A buffer combination can also be created by combining an acid and its conjugate base. An acetate buffer, for example, can be prepared from acetic acid and sodium acetate. A mixture of the base and its conjugate acid can also be used to make an alkaline buffer.

7. What are monoprotic acids?

Make a note of the equilibrium expression first. This demonstrates that when the acid dissociates, it produces an equal quantity of hydrogen ions and anion. An ICE table can be used to compute the equilibrium concentrations of these three components (ICE stands for "initial, change, equilibrium"). 

The initial circumstances are listed in the first row, labelled I: the acid concentration is C0, originally undissociated, so A and H+ concentrations are zero; y is the initial concentration of added strong acid, such as hydrochloric acid. When a strong alkali, such as sodium hydroxide, is introduced, y becomes negative because the alkali eliminates hydrogen ions from the solution. The changes that occur when the acid dissociates are specified in the second row, which is labelled C for "change." The acid concentration falls by x units, but the concentrations of A and H+ both rise by x units. The Vedantu app and website offers free study materials.

What is Buffer in Chemistry?

A solution whose pH is not altered to any great extent by the addition of small quantities of either an acid or base is called buffer solution.

Buffer is also defined as the solution of reserve acidity or alkalinity which resists change of pH upon the addition of a small amount of acid or alkali.

Many chemical reactions are carried out at a constant pH. In nature, there are many systems that use buffering for pH regulation. For example, the bicarbonate buffering system is used to regulate the pH of blood, and bicarbonate also acts as a buffer in the ocean.

Table of Contents

Recommended videos, characteristics of buffer solution, types of buffer solutions.

• Buffer Action

Hendersion’s Equation (pH of buffer)

Buffer capacity, applications of buffer in chemistry.

(i) It has a definite pH.

(ii) Its pH does not change on standing for long periods of time.

(iii) Its pH does not change on dilution.

(iv) Its pH is slightly changed by the addition of small quantity of an acid or base.

(a) Acidic Buffer:

It is formed by the mixture of weak acid and its salt with a strong base.

Examples: (i) CH 3 COOH + CH 3 COONa, (ii) HCN + NaCN, (iii) Boric acid + Borax etc.

(b) Basic Buffer:

It is formed by the mixture of a weak base and its salt with strong acid.

Examples: (i) NH 4 OH + NH 4 Cl, (ii) NH 4 OH + NH 4 NO3, (iii) Glycine + Glycine hydrochloride

(c) Simple Buffer:

It is formed by a mixture of acid salt and normal salt of a polybasic acid,

example Na 2 HPO 4 + Na 3 PO 4

Or a salt of weak acid and a weak base. Example: CH 3 COONH 4

Buffer Actions

It is the mixture of CH 3 COOH and CH 3 COONa in aqueous solution.

CH 3 COOH ⇋ CH 3 COO – + H + (incomplete dissociation)

CH 3 COONa → CH 3 COO – + Na + (complete dissociation)

H 2 O ⇋ H + + OH – (incomplete dissociation)

Action of acid: when a drop of stong acid (HCl) is added in the above buffer solution H+ ions combine with CH3COO- ions to form feebly ionised CH3COOH. Whose ionisation is further suppressed due to common ion effect. So pH of the solution unaltered.

Action of base: when a drop of strong base (NaOH) is added to the above buffer solution it react with free acid to form undissociated water molecules. So pH of the solution unaltered.

CH 3 COOH + OH – ⇋ CH 3 COO – + H 2 O

It is the mixture of NH 4 OH and NH 4 Cl in aqueous solution.

NH 4 OH ⇋ NH 4 + + OH – (incomplete dissociation)

NH 4 Cl → NH 4 + + Cl – (complete dissociation)

Action of acid: when a drop of HCl is added, the added H+ ions combine with NH4OH to form undissociated water molecules. So the pH of buffer is unaffected.

NH 4 OH + OH –   ⇋  NH 4 + + H 2 O

Action of base: when a drop of NaOH is added, the added OH – ions combine with NH 4 + ions to form feebly ionised NH 4 OH. It is further suppressed due to common ion effect. So the pH of buffer is unaffected.

It is a mixture of CH 3 COOH and CH 3 COONa

CH 3 COOH ⇋ CH 3 COO – + H +

CH 3 COONa → CH 3 COO – +  Na +

Taking negative log both sides, we obtain that

– log[H + ] = – log K a – log {[CH 3 COOH]/[CH 3 COO – ]}

pH = pK a + log {[CH 3 COO – ]/[CH 3 COOH]}

pH = pK a + log {[salt] / [acid]}

This equation is known as Hendersion’s Equation

Where, K a = dissociation constant

It is a mixture of NH 4 OH and NH 4 Cl

NH 4 OH ⇋ NH 4 + + OH –

NH 4 Cl → NH 4 + + Cl –

– log [OH – ] = – log K b – log {[NH 4 OH] / [NH 4 + ]}

pOH = pK b + log { [NH 4 + ] / [NH 4 OH]}

pOH = pK b + log {[salt] / [base]}

Where, K b = dissociation constant

pH + pOH = 14

Buffer capacity is defined as the number of moles of acid or base added in one litre of solution as to change the pH by unity.

Buffer capacity (Φ) = No. of moles of acid or base added to 1 litre solution/change in pH

Φ = ∂b /∂(pH)

Where ∂b – No. of moles of acid or base added to 1 litre

∂(pH) – change in pH

(i) Buffers are used in industrial processes such as manufacture of paper, dyes, inks, paints, drugs, etc.

(ii) Buffers are also employed in agriculture, dairy products and preservation of various types of foods and fruits.

(iii) It is used to determine the pH with the help of indicators.

(iv) Blood is the natural buffer, it maintenance of pH is essential to sustain life because enzyme catalysis is pH sensitive process. The normal pH of blood plasma is 7.4.

(v) For the removal of phosphate ion in the qualitative inorganic analysis after the second group using CH 3 COOH + CH 3 COONa buffer.

Frequently Asked Questions – FAQs

What is a buffer and its types.

The solution which opposes the change in their pH value on the addition of small amount of strong acid or strong base is known as buffer solution. These are mainly acidic buffers and basic buffers.

What are the properties of buffers?

The properties of buffet are (i) pH of buffer solution are reserved. (ii) Its pH does not change on standing for long periods of time. (iii) Its pH does not change on dilution.

What is pH stand for?

The abbreviation pH stands for potential hydrogen. It is a scale used to specify the acidity or basicity of an aqueous solution. pH is the negative of the base 10 logarithm of the activity of the H + ion. Mathematically pH = – log [H + ]

What is the pH of blood?

Blood has a normal pH range of 7.35 to 7.45. This means that blood is naturally slightly alkaline or basic.

Is milk alkaline or acid?

The pH of milk is 6.7 to 6.9, making it slightly below neutral and therefore acid-forming. The exception is raw milk, which may be more alkalizing than pasteurized milk.

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5.1: Day 36- Buffer Solutions

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• Page ID 371705

• John Moore, Jia Zhou, and Etienne Garand
• University of Wisconsin

Day 36: Buffer Solutions

D36.1 buffer solutions.

A mixture of a weak acid and its conjugate base, such as acetic acid and sodium acetate (CH 3 COOH + CH 3 COONa), or a mixture of a weak base and its conjugate acid, such as ammonia and ammonium chloride (NH 3 + NH 4 Cl), is a buffer solution. A buffer solution resists changes in pH when small amounts of a strong acid or a strong base are added (Figure 1).

A solution of equal concentrations of CH 3 COOH and CH 3 COONa is slightly acidic because the K a,acetic acid > K b, acetate anion . When a strong base, such as NaOH, is added to this solution, the OH – anions react with the few H 3 O + cations, decreasing concentrations of H 3 O + . This shifts the following equilibrium to the right:

CH 3 COOH(aq) + H 2 O( l ) ⇌ CH 3 COO – (aq) + H 3 O + (aq)

restoring H 3 O + concentration to almost the value it had before the NaOH was added. The net effect of most of the added NaOH is to convert some of the weak acid, CH 3 COOH, to a weak base, CH 3 COO − :

CH 3 COOH(aq) + OH – (aq) → CH 3 COO – (aq) + H 2 O( l )

Hence, there is only a minimal decrease in H 3 O + concentration.

When a strong acid, such as HCl, is added, the net effect of most of the added H 3 O + is to convert acetate anions to acetic acid molecules:

CH 3 COO – (aq) + H 3 O + (aq) → CH 3 COOH(aq) + H 2 O( l )

And again, there is only a minimal increase in H 3 O + concentration.

As illustrated in Figure 2, a buffer solution can moderate changes to pH because it consists of a weak acid that can react with added strong base as well as a weak base that can react with added strong acid.

The weak base and weak acid in a buffer solution are typically a conjugate acid-base pair, which maintains one dynamic equilibrium that responds to additions of other acids and bases. If they are not a conjugate acid-base pair, then there would be two dynamic equilibria at play, which significantly complicates the buffering actions.

Activity 1: pH of a Buffer Solution

Query \(\PageIndex{1}\)

Exercise 1: characteristics of buffer solutions, d36.2 henderson-hasselbalch equation.

The ionization constant expression for a weak acid HA is:

\[K_a = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{A}^{-}]_e}{[\text{HA}]_e} \nonumber \]

Rearranging gives:

\[[\text{H}_3\text{O}^{+}]_e = K_a\;\times\;\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \nonumber \]

Taking the negative logarithm of both sides, we have:

\[\begin{array}{rcl} -\text{log}[\text{H}_3\text{O}^{+}]_e &=& -\text{log}\;K_a\;-\;\text{log}\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \\[1 em] \text{pH} &=& \text{p}K_a\;-\;\text{log}\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \\[1 em] \text{pH} &=& \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_e}{[\text{HA}]_e} \end{array} \nonumber \]

It is much more convenient to deal with the initial concentrations of the weak acid and weak base when preparing a buffer solution. (The initial concentration is the amount of weak acid and weak base added to the solution mixture divide by the volume.) Therefore:

\[\text{pH} = \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_0\;+\;x}{[\text{HA}]_0\;-\;x} \nonumber \]

“ x” is the increase in concentration of H 3 O + as the solution reaches equilibrium (see activity 1 above) and [HA] 0 and [A − ] 0 are the concentration of HA and A − before any reaction occurs. When the approximation that x is at least 100 times smaller than the concentrations of HA and A − is valid, we have the Henderson-Hasselbalch equation :

\[\text{pH} = \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_0}{[\text{HA}]_0} \nonumber \]

Note that when [A−] 0 = [HA] 0 , pH = p K a + log(1) = p K a .

The Henderson-Hasselbalch equation can be used to calculate the buffer solution pH, given the K a and the initial concentrations, or it can be used to determine the ratio of initial concentrations of weak acid and base required to achieve a desired pH.

The Henderson-Hasselbalch equation applies only to buffer solutions in which the ratio

\[\dfrac{[\text{A}^{-}]_0}{[\text{HA}_0} \nonumber \]

is between 0.1 and 10. If enough strong acid or strong base is added to the buffer solution to exceed this range, the pH begins to change significantly (in other words, the solution is no longer a buffer solution).

Exercise 2: Using the Henderson-Hasselbalch Equation to Calculate pH

D36.3 selection of a suitable buffer.

A buffer solution moderates changes in pH because it contains both a weak acid that can react with added strong base and a weak base that can react with added strong acid. This leads to several criteria for selecting a suitable buffer solution for a given purpose.

• The pK a of the weak acid in the buffer should be close to the desired pH of the buffer solution. According to the Henderson-Hasselbalch equation, if the concentrations of weak acid and weak base are equal, the pH of the buffer solution equals the p K a of the weak acid involved.
• A buffer solution should have approximately equal concentrations of the weak acid and weak base. A

ratio of >10 or <0.1 makes for a poor buffer solution. Figure 3 shows how the pH of an acetic acid-acetate ion buffer increases as strong base is added. The initial pH is p K a = 4.74. When pH reaches 5.74, a change of 1 pH unit, the ratio

\[\dfrac{[\text{acetic acid}]}{[\text{acetate anion}]} = 0.11 = 11\text{%} \nonumber \]

After that the pH increases more rapidly and the solution no longer provides significant buffering.

• The larger the amounts (mol) of weak acid and weak base are the greater is the amount (mol) of strong base or strong acid that can be added before there is a significant change in pH . 

When designing a buffer system, look for weak conjugate acid-base pairs that have pK a of the weak acid near the desired pH. Then adjust the ratio of the weak base to weak acid concentrations to achieve the exact pH desired. Make certain that the concentrations of weak base and weak acid are large enough to react with the quantities of acid or base that might be added to the buffer solution. 

Activity 2: Preparing a Buffer Solution with a Desired pH

Query \(\PageIndex{4}\)

D36.4 buffer capacity .

We can see how a buffer solution works by comparing quantitatively the pH of a buffered solution with the pH of a unbuffered solution upon addition of a strong acid or base. 

Activity 3: Calculating pH Change for a Buffer Solution

Query \(\PageIndex{5}\)

Activity 4: pH Change in an Unbuffered Solution

Query \(\PageIndex{6}\)

We can see from the above activities that the change in pH is much more significant in the unbuffered solution compared to the buffered solution.

However, buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 4). For example, if we add sufficient strong base to a buffer that all the weak acid has reacted, no more buffering action toward the base is possible. Similarly, if we add an excess of strong acid, the weak base would all be reacted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to react away all the weak acid or base in a buffer to make significant change in pH: buffering action diminishes rapidly as a given component nears depletion. This was seen in Figure 3 in the above section, where reducing the concentration of weak acid to 11% of the concentration of weak base caused a change of 1 pH unit. The curve in Figure 3 goes up rapidly after that, indicating that the buffer has been “broken” and no longer resists changes in pH.

The buffer capacity is the amount (mol) of acid or base that can be added to a given volume of a buffer solution before the pH changes by ±1 from the pK a of the weak acid. (Recall that if equal concentrations of weak acid and conjugate base are in a buffer solution, pH = pK a .)

Buffer capacity depends on the amount (mol) of weak acid and its conjugate base that are in a buffer mixture. For example, a 1 L solution of 1.0 M CH 3 COOH and 1.0 M CH 3 COONa has a greater buffer capacity than a 1 L solution of 0.10 M CH 3 COOH and 0.10 M CH 3 COONa, even though both solutions have the same pH. The first solution has more buffer capacity because it contains more moles of acetic acid and acetate ion.

It takes 0.82 mol HCl to change the buffer pH from 4.74 to 3.74 in the first solution:

\[\begin{array}{rcl} 3.74 &=& 4.74\;+\;\text{log}\dfrac{[\text{CH}_3\text{COO}^-]_0}{[\text{CH}_3\text{COOH}]_0}\\[0.5em] 10^{3.74-4.74} &=& \dfrac{[\text{CH}_3\text{COO}^-]_0}{[\text{CH}_3\text{COOH}]_0} \\[0.5em] 0.10 &=& \dfrac{(1.0\;\text{mol}\;-\;x)/(1\;\text{L})}{(1.0\;\text{mol}\;+\;x)/(1\;\text{L})} \\[0.5em] 0.10(1.0\;\text{mol}\;+\;x) &=& 1.0\;\text{mol}\;-\;x \\[0.5em] 1.1x &=& 0.90\\[0.5em] x &=& 0.82\;\text{mol} \end{array} \nonumber \]

On the other hand, for the solution where the concentrations of weak acid and conjugate base are 0.10 M, it takes only one-tenth as much HCl, 0.082 mol HCl, to change the buffer pH from 4.74 to 3.74:

\[\dfrac{[\text{CH}_3\text{COO}^-]_0}{[\text{CH}_3\text{COOH}]_0}\;=\;10^{-1}\;=\;0.1\;=\;\dfrac{0.1\;\text{mol}\;-\;x}{0.1\;\text{mol}\;+\;x}\;\;\;\;\;x\;=\;0.082\;\text{mol} \nonumber \]

If a buffer solution does not have equal concentrations of weak acid and weak base, the buffer capacity when strong acid is added is different from the buffer capacity when strong base is added.

Exercise 3: Calculating Buffer Capacity

Podia Question

A solution is prepared by adding 50.0 mL 0.50-M acetic acid and 50.0 mL 0.30-M NaOH to a beaker and stirring. Is this solution a buffer solution? If so, calculate the pH of the buffer. If not, explain why the solution does not resist change in pH when 1.0 mL 0.5-M HCl is added.

Query \(\PageIndex{8}\)

Two days before the next whole-class session, this Podia question will become live on Podia , where you can submit your answer.

Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments.

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Buffer Calculator

About biological buffers.

Buffers are used in many types of scientific studies to maintain or control the acidity of a solution within a desired physiological range, or even to initiate a particular reaction. The pH of substances affects both the rate and efficiency of chemical reactions, and the recovery and purity of products. Hence, buffers considerably contribute to variations in the outcome, and ultimately the success or failure, of studies. Therefore, understanding buffers and error-free calculation of buffers is essential to reduce variability and undesired interactions for reliable research.

Find your pH balance with our biological buffer systems .

Using the buffer preparation calculator

This buffer calculator provides an easy-to-use tool to calculate buffer molarity and prepare buffer solutions using the formula weight of the reagent and your desired volume (L, mL, or µL) and concentration (M, mM, or nM).

To calculate the amount of buffer needed, please select a buffer from the Selection menu. The empirical formula, pKa, buffer pH range, formula weight and product list will appear. Enter the desired final volume and concentration and click “Calculate Mass.” The exact mass of the buffer will be calculated in grams and a step-by-step buffer recipe is automatically provided to assist in buffer preparation.

Click on the list of product results for detailed information on our available buffers and to place an order.

Results based on your selection:

Your stock solution of is calculated to be based on the formula weight of g/mol.

To make a solution, add g of to deionized water.

Adjust the final volume of the solution to with deionized water.

Adjust the pH as needed

Buffer Calculations: Formula and Equations

• Molar solution equation:  desired molarity × formula weight × solution final volume (L) = grams needed
• Percentage by weight (w/v):  (% buffer desired / 100) × final buffer volume (mL) = g of starting material needed
• Henderson-Hasselbach equation: pH = pKa + log [A-]/[HA]

The Henderson-Hasselbalch equation enables determination of a buffer solution's pH when the pKa is known. 1 A buffer solution consists of an acid and a salt of the conjugate base of the acid. If the pH and pKa are known, the amount of salt (A-) and acid (HA) can be calculated. When the amount of salt is equal to the acid, the pH is equal to pKa and max buffering capacity is obtained. 

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